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Difference between revisions of "2020 AMC 10A Problems/Problem 21"

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Note that <math>2^{17}-1</math> equals <math>2^{16}+2^{15}+...+1</math>, since the sum of a geometric sequence is <math>\frac{a^n-1}{a-1}</math>.
 
Note that <math>2^{17}-1</math> equals <math>2^{16}+2^{15}+...+1</math>, since the sum of a geometric sequence is <math>\frac{a^n-1}{a-1}</math>.
 
Thus, we can see that <math>a^{16}-a^{15}</math> forms the sum of 17 different powers of 2.  
 
Thus, we can see that <math>a^{16}-a^{15}</math> forms the sum of 17 different powers of 2.  
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Applying the same thing to each of <math>a^{14}-a^{13}</math>, <math>a^{12}-a^{11}</math>, ... , <math>a^{2}-a^{1}</math>, each of the pairs forms the sum of 17 different powers of 2. This gives us <math>17*8=136</math>.
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Applying the same method to each of <math>a^{14}-a^{13}</math>, <math>a^{12}-a^{11}</math>, ... , <math>a^{2}-a^{1}</math>, each of the pairs forms the sum of 17 different powers of 2. This gives us <math>17*8=136</math>.
 
Finally, we must count the <math>a^0</math> term.  
 
Finally, we must count the <math>a^0</math> term.  
 
Our answer is <math>136+1=\boxed{\textbf{(C) } 137}</math>.
 
Our answer is <math>136+1=\boxed{\textbf{(C) } 137}</math>.

Revision as of 22:28, 31 January 2020

There exists a unique strictly increasing sequence of nonnegative integers $a_1 < a_2 < … < a_k$ such that\[\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.\]What is $k?$

$\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306$

Solution

First, substitute $2^{17}$ with $a$. Then, the given equation becomes $\frac{a^{17}+1}{a+1}=a^{16}-a^{15}+a^{14}...-a^1+a^0$. Now consider only $a^{16}-a^{15}$. This equals $a^{15}(a-1)=a^{15}*(2^{17}-1)$. Note that $2^{17}-1$ equals $2^{16}+2^{15}+...+1$, since the sum of a geometric sequence is $\frac{a^n-1}{a-1}$. Thus, we can see that $a^{16}-a^{15}$ forms the sum of 17 different powers of 2. Applying the same method to each of $a^{14}-a^{13}$, $a^{12}-a^{11}$, ... , $a^{2}-a^{1}$, each of the pairs forms the sum of 17 different powers of 2. This gives us $17*8=136$. Finally, we must count the $a^0$ term. Our answer is $136+1=\boxed{\textbf{(C) } 137}$.

~seanyoon777

Video Solution

https://youtu.be/Ozp3k2464u4

~IceMatrix

See Also

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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